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Differentiation and mathematical induction?
i) Let f : (0,∞) -> R defined by f(x) = (sinx)/(x^1/2). Show that
f ''(x) + (1/x)f '(x) + (1-(1/(4x^2)))f(x) = 0
ii) Define f : R->R by
f(x):= 0, when x ≤ 0 and f(x):= x^(n+1), when x > 0. where n ϵ N.
Prove that f is n-times differentiable on R and find f^(k) for k = 1,....,n.
iii) Use Mathematical induction to prove that
(d^n/dx^n) e^(-x^2) = F_n (x) e^(-x^2)
where F_n is a polynomial of degree n. (Note: you are not required to find F_n)
P.s. you do NOT need to answer all of them :)
1 Answer
- Elizabeth MLv 71 decade agoFavorite Answer
(i) x^(1/2)f(x)=sinx and differentiate, x^(1/2)f'(x)+(1/2)x^(-1/2)f(x)=cosx
Differentiate again:
x^(1/2)f"(x)+(1/2)x^(-1/2)f'(x)-(1/4)x^(-3/2)f(x)+(1/2)x^(-1/2)f'(x)=-sinx=-x^(1/2)f(x)
x^(1/2)f"(x)+x^(-1/2)f'(x)+(-1/(4x^2+1)f(x)=0
(ii) Difficult to know what is expected but I suspect you need to show
that f'(x)=(m+1)x^m, where x>0 for f(x)=x^(m+1), using
f'(x)=Lim h->0 of (f(x+h)-f(x))/h which can be done fairly easily
Then you can use proof by induction to prove the general case.
(iii) Assume true for n=k, I will use D^n for d^n/dx^n.
So that D^k(e^(-x^2))=F_k(x)e^(-x^2)
So D^(k+1)=F'_k(x)e^(-x^2)+F_k(x).-2xe^(x^2)
=e^(-x^2)(F'_k(x)-2xF_k(x)) but 2xF_k(x) is a polynomial in x of degree k+1
an so therefore is (F'_k(x)-2xF_k(x)), giving D^(k+1)=F_k+1(x)e^(-x^2)
So it is true for n+1.
You need to show it is true for k=1, i.e. D(e^(-x^2)) fits formula. I am sure
you can finish.
